Fock state and corresponding relations for continuous momentum label In Wikipedia I found following relation for Fock state:
$$
\hat {a}_i| \{n_j\}_j\rangle ~=~ \sqrt{n}_i| \{n_j-\delta_{ij}\}_j\rangle,
$$
where $n_j$ refers to the number of $j$'th particles. This relation is correct if $$[\hat {a}_{i}, \hat {a}^{+}_{j}]_{\pm} ~=~ \delta_{ij}.$$ But how to modify it in a case of $$[\hat {a}(\mathbf p ), \hat {a}^{+}(\mathbf k)]_{\pm} = \delta (\mathbf p - \mathbf k)?$$ For example, by using
$$
| n + 1\rangle = \frac{\prod_{i}\hat {a}^{+}(\mathbf p_{i})}{\sqrt{(n + 1)!}}| \rangle
$$
and acting by $\hat {a}(\mathbf p_{n + 1})$ on it I can get
$$
\hat {a}| n + 1\rangle = \frac{1}{\sqrt{n + 1}}\delta (\mathbf p - \mathbf k)| n\rangle .
$$
 A: In the context of, for example, a single massive, real scalar field in relativistic quantum field theory, the operators $a(\mathbf k)$ and $a^\dagger(\mathbf k)$ create and annihilate quanta with momentum $\mathbf k$.  
Let's follow the notation of Eric D'Hoker given in his lecture notes.  See especially section 4.1.  In particular, we use the commutation relations between creation and annihilation operators:
\begin{align}
  [a(\mathbf k), a^\dagger(\mathbf k')] &= 2\pi^3\omega_\mathbf k\delta^{(3)}(\mathbf k - \mathbf k'), \qquad \omega_\mathbf k = \sqrt{\mathbf k^2 + m^2} \\
[a(\mathbf k), a(\mathbf k')]&=0 \\
[a^\dagger(\mathbf k), a^\dagger(\mathbf k')] &= 0
\end{align}
If $|0\rangle$ denotes the vacuum, then $a(\mathbf k)|0\rangle = 0$, namely the vacuum is annihilated by destruction operators.  We can then define a state containing a quantum of momentum $\mathbf k$ by applying the creation operator corresponding to that momentum to the vacuum;
\begin{align}
  |\mathbf k\rangle= a^\dagger(\mathbf k)|0\rangle
\end{align}
We can also generate states with $n$ quantua by applying a sequence of $n$ creation operators to the vacuum
\begin{align}
  |\mathbf k_1, \mathbf k_2, \dots, \mathbf k_n\rangle = a^\dagger(\mathbf k_n)\cdots a^\dagger(\mathbf k_2) a^\dagger(\mathbf k_1)|0\rangle
\end{align}
If we apply an annihilation operator to a single-quantum state, then we get
\begin{align}
  a(\mathbf k')|\mathbf k\rangle
&=a(\mathbf k')a^\dagger(\mathbf k)|0\rangle \\
&= \Big(a^\dagger(\mathbf k)a(\mathbf k')|0\rangle-[a^\dagger(\mathbf k),a(\mathbf k')]\Big)|0\rangle \\
&= 2\pi^3\omega_\mathbf k \delta^{(3)}(\mathbf k' - \mathbf k)|0\rangle
\end{align}
and similarly for multi-quantum states.
